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General Physics (PHY 2140)

Lecture 9Lecture 9 Electrodynamics

Electric current temperature variation of resistance electrical energy and power

Chpter 17-18

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Lightning ReviewLightning Review

Last lecture:

1.1. Current and resistanceCurrent and resistance

Current and drift speedCurrent and drift speed Resistance and Ohm’s lawResistance and Ohm’s law

• I is proportional to VI is proportional to V Resistivity Resistivity

• material propertymaterial property

Review Problem: Consider two resistors wired one after another. If there is an electric current moving through the combination, the current in the second resistor is

a. equal tob. halfc. smaller, but not necessarily half the current through the first

resistor.

R Al

QIt

dI nqv A

V IR

a

b

c

R1

R2

I

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17.4 Resistivity - Example(a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321 m.

Cross section: 22 3 7 20.321 10 3.24 10A r m m

Resistivity (Table): 1.5 x 10 m. 6

7 2

1.5 10 4.63.24 10 m

R ml A m

Resistance/unit length:

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17.4 Resistivity - Example(b) If a potential difference of 10.0 V is maintained across a 1.0-m length of the nichrome wire, what is the current?

10.0 2.24.6

V VI AR

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17.4 Temperature Variation of Resistance - Intro• The resistivity of a metal depends on many

(environmental) factors.• The most important factor is the temperature.• For most metals, the resistivity increases with

increasing temperature.• The increased resistivity arises because of larger

friction caused by the more violent motion of the atoms of the metal.

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For most metals, resistivity increases For most metals, resistivity increases approx. linearly with temperature.approx. linearly with temperature.

1o oT T

• is the resistivity at temperature T (measured in Celsius).

• is the reference resistivity at the reference temperature T (usually taken to be 20 oC).

• is a parameter called temperature coefficient of resistivity.

For a conductor with fixed cross section.For a conductor with fixed cross section.

1o oR R T T

TMetallic Conductor

TSuperconductor

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17.5 Temperature Variation of Resistance - 17.5 Temperature Variation of Resistance - ExampleExamplePlatinum Resistance ThermometerPlatinum Resistance ThermometerA resistance thermometer, which measures temperature by measuring the A resistance thermometer, which measures temperature by measuring the change in the resistance of a conductor, is made of platinum and has a change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 resistance of 50.0 at 20 at 20ooC. When the device is immersed in a vessel C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 containing melting indium, its resistance increases to 76.8 . Find the melting . Find the melting point of Indium.point of Indium.

Solution:Using =3.92x10-3(oC)-1 from table 17.1.

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Platinum Resistance ThermometerPlatinum Resistance ThermometerA resistance thermometer, which measures temperature by measuring the change in the A resistance thermometer, which measures temperature by measuring the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 resistance of a conductor, is made of platinum and has a resistance of 50.0 at 20 at 20ooC. C. When the device is immersed in a vessel containing melting indium, its resistance When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 increases to 76.8 . Find the melting point of Indium.. Find the melting point of Indium.

Solution:Solution:Using Using =3.92x10=3.92x10-3-3((ooC)C)-1-1 from table 17.1. from table 17.1. RRoo=50.0 =50.0 ..

TToo=20=20ooC.C.

R=76.8 R=76.8 .. 13

76.8 50.0

3.92 10 50.0

137

157

oo

oo

o

o

R RT T

R C

C

T C

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Appendix: SuperconductivityAppendix: Superconductivity

19111911: H. K. Onnes, who had figured out how to : H. K. Onnes, who had figured out how to make liquid helium, used it to cool mercury to 4.2 make liquid helium, used it to cool mercury to 4.2 K and looked at its resistance:K and looked at its resistance:

19571957: Bardeen (: Bardeen (UIUC!UIUC!), Cooper, and Schrieffer (“BCS”) publish theoretical ), Cooper, and Schrieffer (“BCS”) publish theoretical explanation, for which they get the Nobel prize in 1972.explanation, for which they get the Nobel prize in 1972.

It was Bardeen’s It was Bardeen’s secondsecond Nobel prize (1956 – transistor) Nobel prize (1956 – transistor)

–Current can flow, even if E=0.–Current in superconducting rings can flow for years with no decrease!

At low temperatures the resistance of some At low temperatures the resistance of some metalsmetals0, measured to be less than 0, measured to be less than 1010-16-16••ρρconductorconductor (i.e., (i.e., ρρ<<1010-24 -24 ΩΩmm)!)!

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17.7 Electrical energy and power17.7 Electrical energy and power

In any circuit, battery is used to induce electrical currentIn any circuit, battery is used to induce electrical current chemical energychemical energy of the battery is transformed into of the battery is transformed into kinetic energykinetic energy

of mobile charge carriers (electrical energy gain)of mobile charge carriers (electrical energy gain)

Any device that possesses resistance (resistor) present Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heatin the circuit will transform electrical energy into heat

kinetic energykinetic energy of charge carriers is transformed into of charge carriers is transformed into heatheat via via collisions with atoms in a conductor (electrical energy loss)collisions with atoms in a conductor (electrical energy loss)

I

V = IR

+ -

B A

C D

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Electrical energyElectrical energy

Consider circuit on the right in detailConsider circuit on the right in detailAB: charge gains electrical energy AB: charge gains electrical energy form the batteryform the battery

(battery looses chemical energy)(battery looses chemical energy)

CD: electrical energy lost (transferred CD: electrical energy lost (transferred into heat)into heat)Back to A: same potential energy Back to A: same potential energy (zero) as before(zero) as beforeGained electrical energy = lost Gained electrical energy = lost electrical energy on the resistorelectrical energy on the resistor

A

B

D

CE Q V

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PowerPower

Compute rate of energy loss (power dissipated on the Compute rate of energy loss (power dissipated on the resistor)resistor)

Use Ohm’s lawUse Ohm’s law

Units of power: SI: watt Units of power: SI: watt delivered energy: kilowatt-hours delivered energy: kilowatt-hours

E QP V I Vt t

22 V

P I V I RR

3 61 kWh 10 3600 3.60 10W s J

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ExampleExamplePower Transmission linePower Transmission line

A high-voltage transmission line with resistance of 0.31 A high-voltage transmission line with resistance of 0.31 /km carries 1000A , /km carries 1000A , starting at 700 kV, for a distance of 160 km. What is the power loss due to starting at 700 kV, for a distance of 160 km. What is the power loss due to resistance in the wire? resistance in the wire?

Given:

V=700000 V=0.31 /km L=160 kmI=1000 A

Find:P=?

Observations: 1. Given resistance/length, compute total resistance2. Given resistance and current, compute power loss

0.31 160 49.6R L km km

22 61000 49.6 49.6 10P I R A W

Now compute power

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Mini-quizMini-quiz

Why do the old light bulbs usually fail just after you turn Why do the old light bulbs usually fail just after you turn them on?them on?

When the light bulb is off, its filament is cold, so its resistance is large. Once the switch it thrown, current passes through the filament heating it up, thus increasing the resistance,

This leads to decreased amount of power delivered to the light bulb, as

Thus, there is a power spike just after the switch is thrown, which burns the light bulb.

Resume: electrical devices are better be turned off if there is a power loss

1o oR R T T

2P I R

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Direct Current CircuitsDirect Current Circuits

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Steady current (constant in magnitude and direction)• requires a complete circuit• path cannot be only resistance

cannot be only potential drops in direction of current flow

Electromotive Force (EMF)• provides increase in potential • converts some external form of energy into electrical energy

Single emf and a single resistor: emf can be thought of as a “charge pump”

I

V = IR

+ - V = IR =

18.1 Sources of EMF18.1 Sources of EMF

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EMFEMF

Each real battery has some Each real battery has some internal resistanceinternal resistanceAB: potential increases by AB: potential increases by on on the source of EMF, then the source of EMF, then decreases by Ir (because of decreases by Ir (because of the internal resistance)the internal resistance)Thus, terminal voltage on the Thus, terminal voltage on the battery battery V isV is

Note: Note: is the same as the is the same as the terminal voltage when the terminal voltage when the current is zero (open circuit)current is zero (open circuit)

r

R

A

B C

DV Ir E

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EMF (continued)EMF (continued)

Now add a load resistance RNow add a load resistance RSince it is connected by a Since it is connected by a conducting wire to the battery conducting wire to the battery → → terminal voltage is the same as terminal voltage is the same as the potential difference across the the potential difference across the load resistanceload resistance

Thus, the current in the circuit isThus, the current in the circuit is

r

R

A

B C

D

IR r

E

,V Ir IR orIr IR

EE

Power output:

2 2I I r I R E

Note: we’ll assume r negligible unless otherwise is stated

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Measurements

Voltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device.

V

Ammeters measure current through a device by being placed in series with the device.

A

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Direct Current Circuits

Two Basic Principles:Conservation of ChargeConservation of Energy

Resistance Networks

V IR

RVI

ab eq

eqab

Req

Ia

b

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Resistors in seriesConservation of Charge

I = I1 = I2 = I3

Conservation of Energy Vab = V1 + V2 + V3

RV

IV V V

IVI

VI

VI

VI

VI

VI

R R R R

eqab

eq

1 2 3

1 2 3 1

1

2

2

3

3

1 2 3

a

b

I

R1

V1=I1R1

R2

V2=I2R2

R3

V3=I3R3

Voltage Divider:VV

RRab eq

1 1

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Resistors in parallelConservation of Charge

I = I1 + I2 + I3

Conservation of Energy Vab = V1 = V2 = V3

1

1 1 1 1

1 2 3

1 2 3 1

1

2

2

3

3

1 2 3

RI

VI I I

V

IV

IV

IV

IV

IV

IV

R R R R

eq ab ab

ab ab ab

eq

aI

R1

V1=I1R1

R2

V2=I2R2

R3

V3=I3R3

b

Current Divider: II

RR

eq1

1

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R1=4

R2=3 R3=6

=18V

Example: Determine the equivalent resistance of the circuit as shown.Determine the voltage across and current through each resistor.Determine the power dissipated in each resistorDetermine the power delivered by the battery